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I was reading Peskin and Schroeder's quantum field theory and going through the book mathematically. Then I got stuck at one equation.

Consider a single, non-interacting real scalar field. The book shows that

$$⟨0|ϕ(x)|p⟩=e^{ip⋅x}$$

Which can be interpreted as the position space wavefunction of a single particle state with momentum p (page 24)

and $ϕ(x)$ equals $$ϕ(x)=\int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2w_\mathbf{p}}}(a_\mathbf{p}+a^\dagger_\mathbf{p})e^{ip⋅x}$$

and when $ϕ(x)$ acts on $|0⟩$

$$ϕ(x)|0⟩=\int \frac{d^3p}{(2\pi)^3} \frac{1}{2E_p}e^{-ip⋅x}|\mathbf p⟩$$

How can the following be mathematically shown? $$⟨0|ϕ(x)|p⟩=e^{ip⋅x}$$

Qmechanic
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Quintis
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1 Answers1

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Given $$ϕ(x)=\int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2E_\mathbf{p}}}(a_\mathbf{p}+a^\dagger_\mathbf{p})e^{ip⋅x} ~,$$ consider the hermitean dual of your third equation (2.41), $$\langle 0| ϕ(x)=\int \frac{d^3p'}{(2\pi)^3} \frac{1}{2E_{p'}}e^{ip'⋅x} \langle \mathbf p' | ~,$$ so that P&S (2.42) trivially follows, by virtue of the peculiar normalization (2.36), a mainstay convention of QFT, $$⟨0|ϕ(x)|p⟩=\int \frac{d^3p'}{(2\pi)^3} \frac{1}{2E_{p'}}e^{ip'⋅x} \langle \mathbf p' |\mathbf p\rangle \\ =\int \frac{d^3p'}{(2\pi)^3} \frac{1}{2E_{p'}}e^{ip'⋅x} ~~2E_{p'} (2\pi)^3 \delta^{(3)}(\mathbf {p'-p})= e^{ip⋅x}~.$$

Important: note this state is not fully localized in space, as you might well slip into assuming. Dotting it onto itself it yields a sharply peaked function, but not quite a delta function, as detailed in the linked question.

Cosmas Zachos
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